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Sagot :
Answer:
A. 3.3%
Step-by-step explanation:
We use the normal approximation to the binomial to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
250 babies, boys and girls equally as likely:
This means that [tex]n = 250, p = 0.5[/tex].
Mean and standard deviation:
[tex]\mu = E(X) = np = 250*0.5 = 125[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{250*0.5*0.5}[/tex]
Probability that out of 250 babies born, 110 or fewer will be boys?
Using continuity correction, this is P(X < 110 + 0.5) = P(X < 110.5), which is the p-value of Z when X = 110.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{110.5 - 125}{\sqrt{250*0.5*0.5}}[/tex]
[tex]Z = -1.83[/tex]
[tex]Z = -1.83[/tex] has a p-value of 0.033
0.033*100% = 3.3%, so option A.
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